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Basically the circuit below is what my professor gave us and we need to solve for the the voltage across the 8 ohm resistor, the current through the 2 ohm resistor, power absorbed by the 10 ohm resistor, and the power delivered by the 4 a source. I used regular mesh and I got the wrong values(I checked using KCL and KVL). One thing that I think I got wrong is I assumed that I1 = 3A(feedback). I tried to search the web for a solution and I've seen something called "super mesh". Is this helpful? If yes, can anyone assist me on how to work with it?

Other thing is I want to check if my values are correct using MIT's Circuit Sandbox, however I don't know where should I put the ground so that the circuit will simulate in the way I wanted it to be. Every help is very much appreciated Thanks!

enter image description here

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    \$\begingroup\$ Show us what you tried. \$\endgroup\$ – Dave Tweed Aug 31 '14 at 18:32
  • \$\begingroup\$ I1 = 3A, I2 = 4A, I3 = -4A V8 = (I1 - I3) * 8 = 56V I2ohms = I2 = 4A P1ohm = (I1 - I2) ^ 2 * 1 = 1W basically I'm clueless using mesh in this one \$\endgroup\$ – Jansen Lopez Aug 31 '14 at 18:42
  • \$\begingroup\$ I would have put the ground at the negative terminal of the 9V dc source. \$\endgroup\$ – Amit Hasan Aug 31 '14 at 19:12
  • \$\begingroup\$ I1=3 is ok. But you can't just say that I2=4, I3=-4, only that I2-I3=4. \$\endgroup\$ – apalopohapa Aug 31 '14 at 19:36
  • \$\begingroup\$ Full solution is given hope you understand. \$\endgroup\$ – Amit Hasan Sep 1 '14 at 6:40
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You have to solve this problem using super mesh. Mesh analysis is done easily when the network only contains voltage sources and no current sources. However if it does contain current sources then you have two option.

  1. Convert the current source to voltage source using a parallel resistor. Sometimes you may not find a parallel resistor, in that case you can place resistor in parallel which has a resistance greater than any other resistors in the network. This approach won’t give cent percent correct answer but still reasonable. enter image description here

  2. Second method is to use super mesh. In this approach you temporarily replace current sources with open circuit which will give you a bigger mesh (super mesh). Then apply kvl to it. Then apply kcl where necessary. You can work out some example to get how it is done from example problem of your textbook.

To solve this problem using supermesh –

The conventional way to put current is in clockwise direction. So I am putting it in clock wise instead of your direction. We remove the current sources and redraw the circuit. Now we only have one mesh. We call this mesh Supermesh.

enter image description here

Here, I1 = -3A

We apply KVL in this supermesh as below-

-9 - I3 - (I3 - I1)8- (I2 - I1)1-2(I2)=0

Or, I2+ 3 I3 = -12 ……………….. (1)

Look I have removed 3A current source but I1 is still included in KVL but the 4A current source is not taken into account. This is an exceptional (not all supermesh problems are like this) problem due to the fact that the 3A source is out of the supermesh but 4A source is inside the supermesh.

Now we have to choose node to apply KCL in the original circuit. We apply KCL to a node in the branch where the two meshes intersect. That is node a.

I2 + 4 = I3…………………… (2)

Now calculate the values of current using 1 and 2 no. equation.

Result: I1 = -3A, I2 = -6A, I3 = -2A.

Negative currents indicate the actual direction of current is opposite to what we initially assumed.

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You don't need meshes. Just calculate everything one source at a time, while keeping the other sources zeroed out (i.e. a voltage source is replaced with a wire, and a current source is replaced with an open (cut) circuit). Then sum the results from each source.

So in this case, the voltage source would see nothing but four resistors in series, and both current sources would see (different sets of) 2 series resistors connected in parallel with 2 other series resistors. Hopefully this helps you see how to "zero" the sources.

I'm not familiar with MIT's Circuit Sandbox, but it shouldn't matter where you put the ground since you just need to check the current flowing in different places - so just pick a spot and call it ground!

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    \$\begingroup\$ This is applying the superposition theorem, in case OP wants to lookup a more detailed explanation. It only applies to linear systems. \$\endgroup\$ – ACD Sep 2 '14 at 17:58

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